Difference between revisions of "2014 AMC 12B Problems/Problem 10"

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(Solution 3 (Testing Values))
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==Solution 3 (Testing Values)==
 
==Solution 3 (Testing Values)==
We know that since adding a multiple of <math>55</math> will increase the number, we know that <math>a \leq{c}</math>. After testing values for <math>a</math>, <math>b</math>, and <math>c</math>, adding their squares, we see that using non-zero integers as the digits can yield a maximum value of <math>a^2 + b^2 + c^2</math> as 27, specifically the combination <math>(a, b, c)</math> of <math>(1, 1, 5)</math>. However, we see that no multiple of <math>55</math> can be added to <math>115</math> to achieve <math>511</math>. No combination of non-zero integers can create a higher answer, so we must try testing with a zero. Realizing that only <math>b</math> can be zero, we first try to see if <math>(1, 0, 5)</math> (which creates 26 from the expression <math>1^2 + 0^2 + 5^2</math>, answer choice A) can be added to a multiple of <math>55</math> to create <math>501</math>, which we find it cannot. However, we quickly realize that the next consecutive pair <math>(1, 0, 6)</math> does satisfy this condition, and <math>1^2 + 0^2 + 6^2 = \boxed{\textbf{(D)}\ 37}</math>. We come to this quickly after realizing that if the multiple of <math>55</math> was an even multiple, it would have to end in <math>0</math> and thus the <math>c</math> digit would remain unchanged, so it must be an odd multiple, which will carry over the <math>c</math> digit, so <math>c \geq{6}</math>.
+
We know that since adding a multiple of <math>55</math> will increase the number, we know that <math>a \leq{c}</math>, since these values will flip. After testing values for <math>a</math>, <math>b</math>, and <math>c</math>, adding their squares, we see that using non-zero integers as the digits can yield a maximum value of <math>a^2 + b^2 + c^2</math> as 27, specifically the combination <math>(a, b, c)</math> of <math>(1, 1, 5)</math>. However, we see that no multiple of <math>55</math> can be added to <math>115</math> to achieve <math>511</math>. No combination of non-zero integers can create a higher answer, so we must try testing with a zero. Realizing that only <math>b</math> can be zero, we first try to see if <math>(1, 0, 5)</math> (which creates 26 from the expression <math>1^2 + 0^2 + 5^2</math>, answer choice A) can be added to a multiple of <math>55</math> to create <math>501</math>, which we find it cannot. However, we quickly realize that the next consecutive pair <math>(1, 0, 6)</math> does satisfy this condition, and <math>1^2 + 0^2 + 6^2 = \boxed{\textbf{(D)}\ 37}</math>. We come to this quickly after realizing that if the multiple of <math>55</math> was an even multiple, it would have to end in <math>0</math> and thus the <math>c</math> digit would remain unchanged, so it must be an odd multiple, which will carry over the <math>c</math> digit, so <math>c \geq{6}</math>.
  
 
-Solution by Joeya
 
-Solution by Joeya

Revision as of 03:05, 16 January 2021

Problem

Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \geq{1}$ and $a+b+c \leq{7}$. At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$.

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41$

Solution 1

We know that the number of miles she drove is divisible by $5$, so $a$ and $c$ must either be the equal or differ by $5$. We can quickly conclude that the former is impossible, so $a$ and $c$ must be $5$ apart. Because we know that $c > a$ and $a + c \le 7$ and $a \ge 1$, we find that the only possible values for $a$ and $c$ are $1$ and $6$, respectively. Because $a + b + c \le 7$, $b = 0$. Therefore, we have \[a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}\]

Solution 2

Let the number of hours Danica drove be $k$. Then we know that $100a + 10b + c + 55k$ = $100c + 10b + a$. Simplifying, we have $99c - 99a = 55k$, or $9c - 9a = 5k$. Thus, k is divisible by $9$. Because $55 * 18 = 990$, $k$ must be $9$, and therefore $c - a = 5$. Because $a + b + c \leq{7}$ and $a \geq{1}$, $a = 1$, $c = 6$ and $b = 0$, and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$, or $\boxed{D}$.


Solution 3 (Testing Values)

We know that since adding a multiple of $55$ will increase the number, we know that $a \leq{c}$, since these values will flip. After testing values for $a$, $b$, and $c$, adding their squares, we see that using non-zero integers as the digits can yield a maximum value of $a^2 + b^2 + c^2$ as 27, specifically the combination $(a, b, c)$ of $(1, 1, 5)$. However, we see that no multiple of $55$ can be added to $115$ to achieve $511$. No combination of non-zero integers can create a higher answer, so we must try testing with a zero. Realizing that only $b$ can be zero, we first try to see if $(1, 0, 5)$ (which creates 26 from the expression $1^2 + 0^2 + 5^2$, answer choice A) can be added to a multiple of $55$ to create $501$, which we find it cannot. However, we quickly realize that the next consecutive pair $(1, 0, 6)$ does satisfy this condition, and $1^2 + 0^2 + 6^2 = \boxed{\textbf{(D)}\ 37}$. We come to this quickly after realizing that if the multiple of $55$ was an even multiple, it would have to end in $0$ and thus the $c$ digit would remain unchanged, so it must be an odd multiple, which will carry over the $c$ digit, so $c \geq{6}$.

-Solution by Joeya

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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